A beam of light converges to a point $P$ . A concave lens of focal length $16 c m$ is placed in the path of the convergent beam $12 c m$ from $P$ . How far from the lens does the beam converge?

12 c m

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7.5 c m

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36 c m

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48 c m

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Solution

The correct option is D $48 c m$

Lens formula is given by
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow v = \frac{f u}{f + u}$
Here, the rays are initially about to converge at point P. Hence P acts as the virtual object for concave lens.
$u = + 12 c m$ and $f = - 16 c m$
Hence, $f = \frac{(- 16) (12)}{- 16 + 12}$
$= 4 \times 12 = 48 c m$

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A beam of light converges to a point P. A concave lens of focal length 16 cm is placed in the path of the convergent beam 12 cm from P. How far from the lens does the beam converge?

The correct option is D 48 cm Lens formula is given by 1v−1u=1f ⇒v=fuf+u Here, the rays are initially about to converge at point P. Hence P acts as the virtual object for concave lens. u=+12 cm and f=−16 cm Hence, f=(−16)(12)−16+12 =4×12=48 cm

A beam of light converges to a point $P$ . A concave lens of focal length $16 c m$ is placed in the path of the convergent beam $12 c m$ from $P$ . How far from the lens does the beam converge?